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The original prolbem is as in the figure:

Suppose the square has unit side length, find the area of blue region. enter image description here

The exact solution is: $$\begin{aligned}S=&\frac{\pi-\sqrt{7}}{4}+2 \arccos\left(\frac{5\sqrt{2}}{8}\right)-\frac{1}{4}\arccos\left(-\frac{3}{4}\right)\\ \approx& 0.49263564433675266\end{aligned}$$

Since the areas of the inscribed circle and the tangent sections are rational numbers times $\pi$, I guess the exact solution cannot be obtained by the four arithmetic operations between $\pi$ and rational numbers (i.e., one has to use more advanced methods, e.g., integral, cosine theorem of triangles and so on, than primary math).

The problem here therefore can be converted into:
$$2 \arccos\left(\frac{5\sqrt{2}}{8}\right)-\frac{1}{4}\arccos\left(-\frac{3}{4}\right)-\frac{\sqrt{7}}{4}$$ cannot be represented by $\pi$ and rational numbers through addition, subtraction, multiplication and division.

But how to prove it?

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  • $\begingroup$ Exactly what shape is yellow/blue suppose to be? $\endgroup$ – DanielV Oct 28 '14 at 9:44
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enter image description here

In figure 1, G is the intersection of the circles $C_1: x^2 + y^2 = (2s)^2$ and $C_2: (x – s)^2 + (y – s)^2 = s^2$.

For some $k$, $C_3 : C_1 + kC_2 = 0$ is a family of circles passing through G (and G’).

For a suitable k, we can generate the circle $C_3 : (x – 2s)^2 + (y – 2s)^2 = (\sqrt (2)s)^2$, which has center at $C(2s, 2s)$ and radius $= \sqrt (2)s$. G also lies on this circle and therefore $CG = \sqrt (2)s$.

Furthermore, all of the following can then be found:-

(1) area of ⊿AGC (with sides $2s, \sqrt (2)s, 2\sqrt (2)s$);

(2) angles $GCE, GAH, GEC, GEK, GEJ, GAB, BAG$ can all be calculated; and

(3) areas of sectors $ABG$, and $GAH$.

Move on to figure 2. It shows how the purple region can be found.

enter image description here

Therefore, the areas of the two purple colored butterfly-shaped regions can be found. (See figure 3 below.)

enter image description here

Also in figure 3, the area of the two brown colored butterfly regions can be found.

Summing up the different colored butterfly regions will give the blue region as required.

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  • 1
    $\begingroup$ Wow! Wow! Wow!! $\endgroup$ – marty cohen Dec 16 '14 at 5:28

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